# Dedekind-Hasse norm implies principal ideal ring

## Statement

A commutative unital ring that admits a Dedekind-Hasse norm must be a principal ideal ring.

In particular, an integral domain that admits a Dedekind-Hasse norm must be a principal ideal domain.

## Definitions used

### Dedekind-Hasse norm

Further information: Dedekind-Hasse norm

A Dedekind-Hasse norm on a commutative unital ring $R$ is a function $N$ from the nonzero elements of $R$ to the nonnegative integers with the property that for any elements $a,b \in R$ with $b \ne 0$, it is true that either $a \in (b)$ or there exists an element $r$ in the ideal $(a,b)$ such that $N(r) < N(b)$.

### Principal ideal ring

Further information: Principal ideal ring

A commutative unital ring $R$ is termed a principal ideal ring if every ideal of $R$ is principal.

## Proof

Given: A commutative unital ring $R$ with a Dedekind-Hasse norm $N$. An ideal $I$ of $R$.

To prove: There exists $b \in I$ such that $I = (b)$.

Proof: If $I = 0$, we can take $b = 0$, and we are done.

So, suppose $I \ne 0$. Consider the function $N$ on the nonzero elements of $I$. Since $N$ maps to a well-ordered set, there is an element $b \in I \setminus \{ 0 \}$ such that $N(b) \le N(r)$ for all $r \in I \setminus \{ 0 \}$.

Pick any $a \in I$. If $a = 0$, then $a \in (b)$ or there exists $r \in (a,b)$ with $N(r) < N(b)$.

Note that in the latter case, we have $r \in (a,b) \subseteq I$, so $r \in I$, and $N(r) < N(b)$, contradicting the assumption that $b$ has minimum norm among the nonzero elements of $I$. Hence, we have the case $a \in (b)$. Thus, $I \subseteq (b)$. Conversely, we clearly have $(b) \subseteq I$, so $I = (b)$.